What is the tangent of f(x)=9+5x+2xe^x,\at x=0 ?

The tangent of f(x)=9+5x+2xe^x,\at x=0 is y=7x+9

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tangent of f(x)=9+5x+2xe^x,\at x=0

tangent of

f (x) = 9 + 5 x + 2 x e^{x}, at x = 0

y = 7 x + 9

Solution steps

Find the tangent point:

(0, 9)

Find the slope of

f (x) = 9 + 5 x + 2 x e^{x} : \frac{df ( x )}{d x} = 2 (e^{x} x + e^{x}) + 5

Compute the slope of

f (x) = 9 + 5 x + 2 x e^{x}

x = 0 : m = 7

Find the line with slope m=

7

and passing through

(0, 9) : y = 7 x + 9

y = 7 x + 9

\frac{\partial}{\partial y} (2 x y^{4} e^{y} + 2 x y^{3} + y)

f (x) = 19 - x^{2}, at x = 3

x y^{^{'}} - 2 y^{^{'}} - (x - 2) sin (x) + y = 0

\frac{d}{d x} (\frac{- y}{x})

(- 3, 7), (2, - 5)

What is the tangent of f(x)=9+5x+2xe^x,\at x=0 ?
The tangent of f(x)=9+5x+2xe^x,\at x=0 is y=7x+9